Step 1: Understanding the Concept:
This problem involves definite integration of a piece-wise constant function (step function).
The greatest integer function \( [x] \) returns the largest integer \( n \) such that \( n \le x \).
To integrate such a function, we must partition the interval of integration at every point where the value of \( [x] \) changes, which occurs at integer values of \( x \).
The integral of a constant over an interval is simply the constant multiplied by the length of that interval.
Step 2: Key Formula or Approach:
Break the integral \( \int_1^4 \) into \( \int_1^2 + \int_2^3 + \int_3^4 \).
Identify the constant value of \( [x] \) in each sub-interval.
Step 3: Detailed Explanation:
The function is \( \log[x] \).
1. For \( 1 \le x<2 \), \( [x] = 1 \). Thus, \( \log[x] = \log 1 = 0 \).
2. For \( 2 \le x<3 \), \( [x] = 2 \). Thus, \( \log[x] = \log 2 \).
3. For \( 3 \le x<4 \), \( [x] = 3 \). Thus, \( \log[x] = \log 3 \).
Wait, the upper limit is exactly 4. However, in integration, a single point does not affect the value of the integral, so we consider up to 4.
Now integrate:
\[ f(x) = \int_1^2 0 dx + \int_2^3 (\log 2) dx + \int_3^4 (\log 3) dx \]
\[ f(x) = [0]_{1}^2 + [(\log 2)x]_2^3 + [(\log 3)x]_3^4 \]
\[ f(x) = 0 + \log 2(3 - 2) + \log 3(4 - 3) \]
\[ f(x) = 0 + \log 2 + \log 3 \]
Using the property \( \log a + \log b = \log(ab) \):
\[ f(x) = \log(2 \times 3) = \log 6 \]
Step 4: Final Answer:
The value is \( \log 6 \).
This matches Option (D).